Problem: Real numbers $x$ and $y$ satisfy the equation $x^2 + y^2 = 10x - 6y - 34$. What is $x+y$?
Solution: We can write the equation as
\[x^2 - 10x + y^2 + 6y + 34 = 0.\]Completing the square in $x$ and $y,$ we get
\[(x - 5)^2 + (y + 3)^2 = 0.\]Hence, $x = 5$ and $y = -3,$ so $x + y = \boxed{2}.$